What is the area of the shared region in the figure above that is bounded by the $x$-axis and the curve with the equation $y=x\sqrt{1-x^2}$?
This is the problem I was given. I assumed the answer was $0$ - the positive and negative areas should cancel each other out. That answer was incorrect. I then thought the answer might be $\frac13$ and they are only asking for the area above the $x$-axis. That was also incorrect. The answer that was given as correct was: $\frac23$. I assume that is because it is the area of the top - which is $\frac13$ and the area of the bottom also $\frac13$ - which makes $\frac23$. But, why is the answer not zero? Doesn't integration count area under the $x$-axis as negative? Does the wording of the question say otherwise?
$\endgroup$ 14 Answers
$\begingroup$In your question, it asks for the area of the shaded region, and area is always positive.
For integration, it is taught as the area between the curve and the $x$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $x$-axis, and offset by those below $x$-axis).
Note that Area is absolute, but Net Area is relative
$\endgroup$ $\begingroup$The integral $\int_{-1}^0y~dx$ is negative, but the area is always non-negative. So the area of the region is $\int_0^1y~dx-\int_{-1}^0y~dx$.
$\endgroup$ $\begingroup$The function is odd, the area you look for is$$A=2\int_0^1x\sqrt{1-x^2}dx$$
put $y=x^2$.
then
$$A=\int_0^1\sqrt{1-y}dy$$$$=\int_0^1(1-y)^{\frac 12}dy$$
$$=\Bigl[ \frac 23(1-y)^{\frac 32}\Bigr]_1^0$$$$=\frac 23.$$
$\endgroup$ $\begingroup$Let $x=\sin t$.
Thus, we need to get $$2\int\limits_0^1x\sqrt{1-x^2}dx=2\int\limits_0^{\frac{\pi}{2}}\sin{t}\cos^2tdt=\int\limits_0^{\frac{\pi}{2}}\sin2t\cos{t}=$$$$=\frac{1}{2}\int\limits_0^{\frac{\pi}{2}}(\sin3t+\sin{t})dt=-\frac{1}{6}\cos3t-\frac{1}{2}\cos{t}\big{|}_0^{\frac{\pi}{2}}=\frac{2}{3}.$$
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